Question

A proton is initially at rest. After some time, a uniform electric field is turned on and the proton accelerates. The magnitude of the electric field is 1.56 x 105 N/C. (a) What is the speed of the proton after it has travelled 4.00 cm? m/s (b) What is the speed of the proton after it has travelled 40.0 cm? m/s

Answer 1

given Page No Date Ans. uzomis Cat Resty. q = charge op Pooton = 1.6 X10 m= mass of Proton = 1.67 X10 kg E=electric field = 1.56X10 Nc So applied forceon Proton f=gE F = 116 Xiöt9 X 1.56X165 If = 2.496X10 N o a We know that by Newtoo's and Law f = ma a= F - 2.496 x 10 1.67 X1527 a=1.4946 X10 ml 62 by and equation of motion V = V² taas initially at Rest Vo= uzoms V² = 205 - 13 =2x1.4946X10X (distance to relled by Poston s = distance travelled by Photon

- 2 (Date car Page No. for s=4.00cm = 4x10² So 13 2x1.9946 X 10 X 4X10 - 11.9568 X 10 v2 = 119.568x10° V = 119.568 810 m/s Y = 10.938 109 m/s So speed will be 10.934 X (b) for s=youm V² = ads - V2 = 2x1.4946X16°x 40 xjö? V = 11.9568x162 V = 3.45786 X10 so when it travels 40 cm its speed will be 3,45786 810 m/s
Thank you